Loser66
  • Loser66
Let A, B \(\in M(n, \mathbb C)\)and assume B is invertible. Show that 1) \((B^{-1}AB)^k= B^{-1}A^kB\) 2) Use 1 to show \(e^{tB^{-1}AB}=B^{-1}e^{tA}B\) Please help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
1) is easy. But I don't know how to do 2 @ganeshie8 @dan815
ganeshie8
  • ganeshie8
is it given that A is diagonal matrix ?
Loser66
  • Loser66
Nope, that is all.

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More answers

UsukiDoll
  • UsukiDoll
invertible meaning non-singular matrix.. there's an inverse.
Loser66
  • Loser66
nvm, I got it. :) Thanks for being here.
Loser66
  • Loser66
for 1) \((B^{-1}AB)^k = (B^{-1}AB)(B^{-1}AB)........(B^{-1}AB)\) open parenthesis and \(BB^{-1}=I\) gives us the answer
Loser66
  • Loser66
for 2) by definition \(e^{tA}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\)
Loser66
  • Loser66
we replace A by \(B^{-1}AB\) then just simplify it to get the answer.
ganeshie8
  • ganeshie8
\(e^{tB^{-1}AB}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}(B^{-1}AB)^k =\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB \) ?
UsukiDoll
  • UsukiDoll
yeah.... this proof looks familiar. Miracrown helped me with one of them.
ganeshie8
  • ganeshie8
how do you conclude
Loser66
  • Loser66
@ganeshie8 sum and t, k are constants, right? hence, we can distribute inside, between \(B^{-1}\) and A
ganeshie8
  • ganeshie8
Are you saying we can take \(B^{-1}\) out ?
Loser66
  • Loser66
Yes
ganeshie8
  • ganeshie8
\(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB =B^{-1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B \) ? this looks ok intuitively but idk what you're gonna use as justification
Loser66
  • Loser66
Look at the definition of \(e^{tA}\) above
ganeshie8
  • ganeshie8
Yes looked.
Loser66
  • Loser66
And we done, right?
ganeshie8
  • ganeshie8
I am asking the justification for that distribution
ganeshie8
  • ganeshie8
I think its not so obvious, at least for me it is not.
Loser66
  • Loser66
https://en.wikipedia.org/wiki/Matrix_multiplication
dan815
  • dan815
|dw:1437277499951:dw|
dan815
  • dan815
|dw:1437277542960:dw|
dan815
  • dan815
|dw:1437277640041:dw|
dan815
  • dan815
therefore true
dan815
  • dan815
if u expand that summation u can show these 2 summations are equation therefore
dan815
  • dan815
|dw:1437277900211:dw|
Loser66
  • Loser66
hey, barbecue, @ganeshie8 asked whether we can put the sum between B^-1 and A or not. You proved \(A(B_1+B_2) = AB_1 +AB_2\) which is irrelevant to the question. ha!!
dan815
  • dan815
this is what we need
dan815
  • dan815
\[\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB =B^{-1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B\]
dan815
  • dan815
|dw:1437278072521:dw|
dan815
  • dan815
you see?
Loser66
  • Loser66
Oh, yeah, barbecue!! Thanks for making it clear.
dan815
  • dan815
sure thing
anonymous
  • anonymous
consider \(C=B^{-1}AB\). now suppose \(C^{n-1}=B^{-1}A^{n-1}B\). it follows that $$C^n=C\cdot C^{n-1}=B^{-1}ABB^{-1}A^{n-1}B=B^{-1}AA^{n-1}B=B^{-1}A^nB$$. so by induction it follows \(C^n=B^{-1}A^n B\) define \(e^X=\sum_{n=0}^\infty \frac1{n!}X^n\) in analogy with the real exponential. $$\begin{align*}e^{B^{-1}AB}=\sum_{n=0}^\infty\frac1{n!}(B^{-1}AB)&=\sum_{n=0}^\infty\frac1{n!}B^{-1}A^nB\\&=B^{-1}\sum_{n=0}^\infty\frac1{n!}A^n B\\&=B^{-1}\left(\sum_{n=0}^\infty\frac1{n!}A^n\right)B\\&=B^{-1}e^AB\end{align*}$$
anonymous
  • anonymous
the above follows because matrix multiplication distributes over addition: $$B^{-1}(A_1+A_2)B=B^{-1}(A_1B+A_2B)=B^{-1}A_1B+B^{-1}A_2B$$

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